tag:blogger.com,1999:blog-6933544261975483399.post2656189481756125812..comments2024-09-10T22:02:29.582-07:00Comments on GoGeometry.com (Problem Solutions): Geometry Problem 1565: Find the Length of BF in Triangle ABCAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-74766116935656867242024-07-22T11:11:32.494-07:002024-07-22T11:11:32.494-07:00Extend ED to X such that D is the midpoint of EX
...Extend ED to X such that D is the midpoint of EX<br /><br />Then AECX is a parallelogram and AE = CX<br />Hence Triangles ABE & CFX are congruent (Hypotenuse, side, right angle)<br /><br />So BE = FX, hence BF = EX = 10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38842782240340024192024-07-21T15:06:55.044-07:002024-07-21T15:06:55.044-07:00On the line BD take a point G symmetric to B relat...On the line BD take a point G symmetric to B relative to D, so that DG = BD. Then triangle DGC is congruent to ABD (AD = DC, BD = DG, angle <BDA = <CDG, SAS case). Project point C to the line BDG, and call it H. As AD = DC, <EDA = <CDH, <br />so DH = ED = 5. <br />Also HG = BE, because ABE and CGH are right triangles, angle <ABE = <CGH and CG = AB.<br />Now, CF = AB (by hypothesis) and CG = AB (congruence of triangles ABD and DGC as seen before), implies that CF = CG and triangle FCG is isosceles. Then H is the middle point of FG, so that FH = HG, and as we saw HG = BE, and then FH = BE.<br />But,<br />FH= FE + ED + DH = FE + 5 + 5 = FE + 10;<br />BE = BF + FE, <br />So FH = BE implies FE + 10 = BF + FE, and cancelling FE we obtain BF = 10.<br /><br />Joaquim Maia<br />Rio de Janeiro<br />BrazilAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83207596775742380112024-07-21T01:11:39.690-07:002024-07-21T01:11:39.690-07:00Extent BD=DG. Draw CH┴DG (H on DG)
TrABD, TrCGD a...Extent BD=DG. Draw CH┴DG (H on DG)<br />TrABD, TrCGD are congruence (SAS) so GC=AB=CF,<ABD=<G<br />TrEAB,TrHCG are congruence (Right Triangles, CG=AB,<ABE=<G) so<br />BE=HG but HG=HF (TrCGH=TrCFH) so BE=HF<br />BE=HF → BF+EF=EF+EH → BF=EH=5+5=10 (EH=2ED)<br />Georgios Kousinioris, Gastouni Greecenoreply@blogger.com